Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;

typedef long long ll;
const ll MAXN=100005;
const ll INF=0x3f3f3f3f;

ll n,K,tot;
ll sum[MAXN];
ll f[MAXN],l[MAXN],r[MAXN];
ll num[MAXN*2];

struct node{
    ll val,tag;
    node *ch[2];
    void PushDown(){
        if(!tag) return;
        if(ch[0]) ch[0]->val=ch[0]->tag=tag;
        if(ch[1]) ch[1]->val=ch[1]->tag=tag;
        val=tag;
        tag=0;
    }
}tree[MAXN*4];
node *root,*ncnt;

void Init(){ncnt=&tree[0];}

node *newnode(ll val=0){
    node *p=++ncnt;
    p->val=val,p->tag=0;
    p->ch[0]=p->ch[1]=NULL;
    return p;
}

void Update(node *&rt,ll l,ll r,ll st,ll ed,ll val){
    if(st>ed) return;
    if(!rt) rt=newnode();
    if(st<=l&&r<=ed){
        rt->val=rt->tag=val;
        return;
    }
    if(!rt->ch[0]) rt->ch[0]=newnode();
    if(!rt->ch[1]) rt->ch[1]=newnode();
    rt->PushDown();
    ll mid=(l+r)>>1;
    if(st<=mid) Update(rt->ch[0],l,mid,st,ed,val);
    if(ed>mid) Update(rt->ch[1],mid+1,r,st,ed,val);
}

ll Query(node *rt,ll l,ll r,ll pos){
    if(!rt) return 0;
    if(l==r) return rt->val;
    ll mid=(l+r)>>1;
    if(!rt->ch[0]) rt->ch[0]=newnode();
    if(!rt->ch[1]) rt->ch[1]=newnode();
    rt->PushDown();
    if(pos<=mid) return Query(rt->ch[0],l,mid,pos);
    return Query(rt->ch[1],mid+1,r,pos);
}

ll Ask(ll k){
    ll pos=Query(root,1,tot,k);
    return pos?f[pos]+(num[l[pos]]-num[k]+K)%K:0;
}

int main(){
    Init();
    scanf("%lld%lld",&n,&K);
    for(ll i=1;i<=n;i++){
        ll a,b;
        scanf("%lld%lld",&a,&b);
        sum[i]=sum[i-1]+a;
        if(b==2) l[i]=0,r[i]=K-1;
        else l[i]=(K-2ll*sum[i-1]%K)%K,r[i]=(K-2ll*sum[i]%K)%K;
        if(b==1&&a*2ll>K) return puts("-1"),0;
        num[++tot]=l[i];
        num[++tot]=r[i];
    }
    sort(num+1,num+tot+1);
    tot=unique(num+1,num+tot+1)-num-1;
    for(ll i=1;i<=n;i++){
        l[i]=lower_bound(num+1,num+tot+1,l[i])-num;
        r[i]=lower_bound(num+1,num+tot+1,r[i])-num;
    }
    for(ll i=n;i>=1;i--){
        f[i]=Ask(l[i]);
        if(l[i]>r[i]) Update(root,1,tot,r[i]+1,l[i]-1,i);
        else Update(root,1,tot,1,l[i]-1,i),Update(root,1,tot,r[i]+1,tot,i);
    }
    ll ans=f[1];
    for(ll i=tot;i>0;i--) ans=min(ans,Ask(i));
    printf("%lld\n",ans+2ll*sum[n]);
}